Re: Q: About number of primes with n digits?




Danny wrote:
jankrihau@xxxxxxxxxxx wrote:

Danny wrote:
The first 4 primes are single digits in length.
The next 21 primes are 2 digits in length.
The next 143 are 3 digits in length.
etc..

4, 21, 143, 1061, 8363, 68906, 586081, 5096876, 45086079,
404204977, 3663002302, 33489857205, 308457624821, 2858876213963,
26639628671867, 249393770611256, 2344318816620308,
22116397130086627, 209317712988603747, 1986761935284574233,
18906449883457813088, 180340017203297174362

Sequence is in OEIS as A006879.

Will the ratio between terms converge?

If the sequence is divergent then at any point can the next
ratio be < the previous ratio?

Dan

By the PNT, the nth term is asymptotically

0.9 * 10^n / (n log 10)

so the ratio converges to 10.

---
J K Haugland
http://home.no.net/zamunda

My reasoning is probably way off but if the
above formula you give is only asympototically
correct how can the ratio be an absolute convergence
too 10?

I have been looking up the PNT and see nothing
about any value pertaining to the different (n) lengths
of the primes. Only the different methods used for estimating
pi(x) not digit length (n) counting.
Also nothing about ratio convergence of this particular
count.

Then again, I could have over looked something.

Thanks,

Dan

The prime counting function pi(x) is defined to be the
number of (positive) primes less than or equal to x,
for any real number x. The Prime Number Theorem
says simply that pi(x) ~ x/ln(x), but a more precise
statement can be given in terms of the logarithmic
integral, esp. if one assumes the Riemann Hypothesis.

If the asymptotic character of this result bothers you,
a simple restatement is that pi(x)/(x/ln(x)) tends to 1
as x tends to +infinity.

In any case, isn't it evident that the "primes of length
k digits" is exactly:

pi(10^k) - pi(10^(k-1))

for integer k > 1?

regards, chip

.



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